By Newton's second law, the net vertical force on the block is
∑ F = N - mg = 0
so that the normal force exerted by the surface has magnitude
N = mg = (2.52 kg) (9.80 m/s²) = 24.7 N
Then as the block slides over the surface, it feels a frictional force of
f = µ (24.7 N)
where µ is the coefficient of kinetic friction.
As the block is pushed by the spring to its equilibrium position, friction performs
µ (-24.7 N) (0.0973 m) = -2.40µ J
of work (which is negative because it opposes the block's motion).
In compressing the spring by 9.73 cm = 0.0973 m, we store
1/2 (116 N/m) (0.0973 m)² = 0.549 J
of energy. This energy is released and partially converted to kinetic energy, while the rest is lost to friction.
By the work-energy theorem, the total work performed on the block as the spring pushes it towards the equilibrium position is equal to the change in its kinetic energy:
W = ∆K
0.549 J - 2.40µ J = 1/2 (2.52 kg) v ² - 0
where v is the speed of the block at the equilibrium position. Solving for v, we get
v = 0.891 √(0.549 - 2.40µ) m/s
After the block is released, the only force acting on it as it slides freely is friction. It comes to a stop after 0.537 m, so that friction performs
µ (-24.7 N) (0.537 m) = -13.3µ J
of work.
Using the work-energy theorem again, we have
W = ∆K
-13.3µ J = 0 - 1/2 (2.52 kg) v ²
Substitute the velocity we found in terms of µ, and solve for µ :
-13.3µ J = -1/2 (2.52 kg) (0.891 √(0.549 - 2.40µ) m/s)²
===> µ = 0.0350
